Function $f (x) = tan x$ is discontinuous at-

x = 0

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x = π / 2

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x = π

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x = - π

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Solution

The correct option is B $x = π / 2$
$lim x \to 0 tan x = lim x \to π tan x = lim x \to - π tan x = 0$
and $lim x \to \frac{π}{2} tan x = \infty$
Therefore, $tan x$ is discontinuous at $x = \frac{π}{2}$

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f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} x sin x . i f 0 < x \leq \frac{π}{2} \frac{π}{2} sin (π + x), i f \frac{π}{2} < x < π \end{matrix}

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f (x) = \frac{sin x}{x}

(0, \frac{π}{2})

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0 < x < \frac{π}{2}

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